Groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 10 July 2011

Groups and group homomorphisms

• A group is a set $G$ with a function

  $\begin{array}{ccc}G\times G& \to & G\\ (g_1,g_2)& ⟼& g_1{g}_2\end{array}$

  such that

  (a)   If $g_1,g_2,g_3\in R$ then $\left(g_1{g}_2\right)g_3=g_1\left(g_2{g}_3\right)$,

  (b)   There exists an identity, $1\in G$, such that if $g\in G$ then $1g=g1=g$,

  (c)   If $g\in G$ then there exists an inverse to $g$, $g^{-1}\in G$, such that $g^{-1}g=g{g}^{-1}=1$.

HW: Show that the identity $1\in G$ is unique.
HW: Show that if $g\in G$ then the inverse $g^{-1}\in G$ is unique.
HW: Why isn't $\{1,2,3,4,5\}$ a group?

Important examples of groups are:

(a)   The integers $\mathbb{Z}$ with the operation of addition,

(b)   The integers mod $n$, $\mathbb{Z}/n\mathbb{Z}$, with operation addition,

(c)   The symmetric group $S_n$,

(c)   The general linear group of invertible $n\times n$ matrices ${\mathrm{GL}}_n\left(ℂ\right)$.

Group homomorphisms are for comparing groups.

Let $G$ and $H$ be groups with identities $1_G$ and $1_H$, respectively.

• A group homomorphism from $G$ to $H$ is a function $f:G\to H$ such that

  (a)   If $g_1,g_2\in G$ then $f\left(g_1{g}_2\right)=f\left(g_1\right)f\left(g_2\right)$,

  (b)   $f\left(1_G\right)=1_H$.

• A group isomorphism is a bijective group homomorphism.
• Two groups $G$ and $H$ are isomorphic, $G\simeq H$, if there exists a group isomorphism $f:G\to H$ between them.

Two groups are isomorphic if the elements of the rings and their operations match up exactly. Think of groups that are isomorphic as being "the same". When classifying groups we put two groups in the same class only if they are isomorphic. This is what is meant by classifying groups "up to isomorphism".

Let $f:G\to H$ be a group homomorphism. Let $1_G$ and $1_{\mathrm{H}}$ be the identities in $G$ and $H$, respectively. Then

(a)   $f\left(1_G\right)=1_H$.

(b)   If $g\in G$ then $f\left(g^{-1}\right)={f\left(g\right)}^{-1}$.

		Proof.
	(a)   Multiply both sides of the following equation by $f{\left(1_G\right)}^{-1}$. $$f\left(1_G\right)=f\left(1_G·1_G\right)=f\left(1_G\right)f\left(1_G\right).$$ (b)   Since $f\left(g\right)f\left(g^{-1}\right)=f\left(g{g}^{-1}\right)=f\left(1_G\right)=1_H$ and $f\left(g^{-1}\right)f\left(g\right)=f\left(g^{-1}g\right)=f\left(1_G\right)=1_H,$ $${f\left(g\right)}^{-1}=f\left(g^{-1}\right).$$ $\square$

• A subgroup $H$ of a group $G$ is a subset $H\subseteq G$ such that

  (a)   If $h_1,h_2\in H$ then $h_1{h}_2\in H$,

  (b)   $1\in H$,

  (c)   If $h\in H$ then $h^{-1}\in H$.

• The trivial group $\left\{1\right\}$ is the set containing only $1$ with the operation given by $1\cdot 1=1$.

Cosets

Let $G$ be a group and let $H$ be a subgroup of $G$. We will use the subgroup $H$ to divide up the group $G$.

• A left coset of $H$ in $G$ is a set $gH=\left\{gh\right|h\in H\}$, where $g\in G$.
• $G/H$ (pronounced "$G$ mod $H$") is the set of left cosets of $H$ in $G$.
• A right coset of $H$ in $G$ is a set $Hg=\left\{hg\right|h\in H\}$, where $g\in G$.
• $H\backslash G$ is the set of right cosets of $H$ in $G$.

Unless specified otherwise we will always work with left cosets and just call them cosets.

HW: Let $G$ be a group and let $H$ be a subgroup of $G$. Let $x$ and $g$ be two elements of $G$. Show that $x\in gH$ if and only if $gH=xH$.

Let $G$ be a group and let $H$ be a subgroup of $G$. Then the cosets of $H$ in $G$ partition $G$.

		Proof.
	To show: (a)   If $g\in G$ then there exists $g\prime \in G$ such that $g\in g\text{'}H$. $\phantom{\text{To show:}}$ (b)   If $g_{1}H\cap g_{2}H\ne \varnothing$ then $g_{1}H=g_{2}H$. (a)   Let $g\in G$. $\phantom{\text{(a)}}$   Then $g=g·1\in gH$, since $1\in H$. $\phantom{\text{(a)}}$   So $g\in gH$. (b)   Assume $g_{1}H\cap g_{2}H\ne \varnothing$. $\phantom{\text{(b)}}$   To show: (ba)   $g_{1}H\subseteq g_{2}H$. $\phantom{\text{(b)}}$   $\phantom{\text{To show:}}$ (bb)   $g_{2}H\subseteq g_{1}H$. $\phantom{\text{(b)}}$   Let $k\in g_{1}H\cap g_{2}H$. $\phantom{\text{(b)}}$   Suppose $k=g_1{h}_1=g_2{h}_2$, where $h_1,h_2\in H$. $\phantom{\text{(b)}}$   Then $$g_1=g_1{h}_1{h_1}^{-1}=k{h_1}^{-1}=g_2{h}_2{h_1}^{-1},\text{and}{g}_2=g_2{h}_2{h_2}^{-1}=k{h_2}^{-1}=g_1{h}_1{h_2}^{-1}.$$ (ba)   Let $g\in g_{1}H$. $\phantom{\text{(ba)}}$   Then there exists $h\in H$ such that $g=g_{1}h$. $\phantom{\text{(ba)}}$   Then $$g=g_{1}h=g_2{h}_2{h_1}^{-1}h\in g_{2}H,\text{since}{h}_2{h_1}^{-1}h\in H.$$ $\phantom{\text{(ba)}}$   So $g_{1}H\subseteq g_{2}H$. $\phantom{\text{(b)}}$   (bb)   Let $g\in g_{2}H$. $\phantom{\text{(bb)}}$   Then there exists $h\in H$ such that $g=g_{2}h$. $\phantom{\text{(bb)}}$   Then $$g=g_{2}h=g_1{h}_1{h_2}^{-1}h\in g_{1}H,\text{since}{h}_1{h_2}^{-1}h\in H.$$ $\phantom{\text{(bb)}}$   So $g_{2}H\subseteq g_{1}H$. So $g_{1}H=g_{2}H$. So the cosets of $H$ in $G$ partition $G$. □

Let $G$ be a group and let $H$ be a subgroup of $G$. If $g_1,g_2\in G$ then $$\mathrm{Card}\left(g_{1}H\right)=\mathrm{Card}\left(g_{2}H\right).$$

		Proof.
	To show: There is a bijection from $g_{1}H$ to $g_{2}H$. Define a map $\phi$ by $$\begin{array}{rcll}\phi :& g_{1}H& \to & g_{2}H\\ & x& \mapsto & g_2{g_1}^{-1}x.\end{array}$$ To show: (a)   $\phi$ is well defined. $\phantom{\text{To show:}}$ (b)   $\phi$ is a bijection. (a)   To show: (aa)   If $x\in g_{1}H$ then $\phi \left(x\right)\in g_{2}H$. $\phantom{\text{(a) To show:}}$ (ab)   If $x=y$ then $\phi \left(x\right)=\phi \left(y\right)$. (aa)   Assume $x\in g_{1}H$. $\phantom{\text{(aa) }}$ Then $x=g_{1}h$ for some $h\in H$. $\phantom{\text{(aa) }}$ So $\phi \left(x\right)=g_2{g_1}^{-1}{g}_{1}h=g_{2}h\in g_{2}H$. (ab)   This is clear from the definition of $\phi$. So $\phi$ is well defined. (b)   By virtue of Theorem 2.2.3 Part I, ???????????????????? we want to construct an inverse map for $\phi$. $\phantom{\text{(b) }}$ Define $$\begin{array}{rcll}\psi :& g_{2}H& \to & g_{1}H\\ & y& \mapsto & g_1{g_2}^{-1}y.\end{array}$$ $\phantom{\text{(b) }}$ HW: Show (exactly as in (a) above) that $\psi$ is well defined. Then $$\psi \left(\phi \right(x)=g_1{g_2}^{-1}\phi (x)=g_1{g_2}^{-1}{g}_2{g_1}^{-1}x=x,\text{and}$$ $$\phi \left(\psi \right)y)=g_2{g_1}^{-1}\phi (y)=g_2{g_1}^{-1}{g}_1{g_2}^{-1}y=y.\phantom{\text{and}}$$ $\phantom{\text{(b) }}$ So $\psi$ is an inverse function to $\phi$. So $\phi$ is a bijection. $\square$

Let $H$ be a subgroup of a group $G$. Then $$\mathrm{Card}\left(G\right)=\mathrm{Card}\left(G/H\right)\mathrm{Card}\left(H\right).$$

		Proof.
	By Proposition 2.2, all cosets in $G/H$ are the same size as $H$. Since the cosets of $H$ partition $G$, the cosets are disjoint subsets of $G$, and $G$ is a union of those subsets. So $G$ is the union of $\mathrm{Card}\left(G/H\right)$ disjoint subsets all of which have size $\mathrm{Card}\left(H\right)$. $\square$

The above results show that the cosets of a subgroup $H$ divide the group $G$ into equal size pieces, one of these pieces being the subgroup $H$ itself.

• A set of coset representatives of $H$ in $G$ is a set of distinct elements $\left\{g_i\right\}$ of $G$ such that
  1. each coset of $H$ is of the form $g_{i}H$ for some $g_i$ and
  2. $g_{i}H\ne g_{j}H$ unless $g_i=g_j$.
• The index of a subgroup $H$ in a group $G$, $\left|G:H\right|$, is the number of cosets of $H$ in $G$, $$\left|G:H\right|=\mathrm{Card}\left(G/H\right).$$

HW:Show that $\left|G:\right\{1\left\}\right|=\mathrm{Card}\left(G\right)$.

Quotient groups $\leftrightarrow$ Normal subgroups

Let $H$ be a subgroup of a group $G$. We can try to make the set of cosets of $H$ into a group by defining an operation on the cosets. The only problem is that this doesn't work for the cosets of just any subgroup, the subgroup has to have special properties.

• A normal subgroup $N$ is a subset of a group $G$ such that if $n\in N$ and $g\in G$ then $gn{g}^{-1}\in N$.

HW: Show that a subgroup $N$ of a group $G$ is normal if and only if $N$ satisfies: if $g\in G$ then $gN=Ng$.

Let $N$ be a subgroup of a group $G$. Then $N$ is a normal subgroup of $G$ if and only if $G/N$ with operation given by $\left(aN\right)\left(bN\right)=\left(abN\right)$ is a group.

		Proof.
	Proof $\Rightarrow$ Assume $N$ is a normal subgroup of $G$. To show: $\left(aN\right)\left(bN\right)=\left(abN\right)$ is a well defined operation on $\left(G/N\right)$. $N$ is the identity element of $G/N$. $g^{-1}N$ is the inverse of $gN$. We want the operation on $G/N$ given by $$\begin{array}{rcl}G/N\times G/N& \to & G/N\\ (aN,bN)& \mapsto & abN\end{array}$$ to be well defined. To show: If $(a_{1}N,b_{1}N),(a_{2}N,b_{2}N)\in G/N\times G/N$ and $(a_{1}N,b_{1}N)=(a_{2}N,b_{2}N)$ then $a_1{b}_{1}N=a_2{b}_{2}N$. Assume $(a_{1}N,a_{2}N),(a_{2}N,b_{2}N)\in G/N\times G/N$ and $(a_{1}N,b_{1}N)=(a_{2}N,b_{2}N)$. Then $a_{1}N=a_{2}N$ and $b_{1}N=b_{2}N$. To show: aa. $a_1{b}_{1}N\subseteq a_2{b}_{2}N.$ $\phantom{\text{To show:}}$ ab. $a_2{b}_{2}N\subseteq a_1{b}_{1}N$. Since $a_{1}N=a_{2}N$, we know $a_1=a_1·1\in a_{2}N$. So there exists $n_1\in N$ such that $a_1=a_2{n}_1$. Similarly, there exists $n_2\in N$ such that $b_1=b_2{n}_2$. Let $k\in a_1{b}_{1}N$. Then there exists $n\in N$ such that $k=a_1{b}_{1}n$. So $$\begin{array}{rcl}k& =& a_1{b}_{1}n\\ & =& a_2{n}_1{b}_2{n}_{2}n\\ & =& a_2{b}_2{b_2}^{-1}{n}_1{b}_2{n}_{2}n.\end{array}$$ Since $N$ is normal, ${b_2}^{-1}{n}_1{b}_2\in N,$ and therefore $\left({b_2}^{-1}{n}_1{b}_2\right)n_{2}n\in N$. So $k=a_2{b}_2\left({b_2}^{-1}{n}_1{b}_2\right)n_{2}n\in a_2{b}_{2}N.$ So $a_1{b}_{1}N\subseteq a_2{b}_{2}N$. To show: $a_2{b}_{2}N\subseteq a_1{b}_{1}N$. Since $a_{1}N=a_{2}N$, there exists $n_1\in N$ such that $a_1{n}_1=a_2$. Since $b_{1}N=b_{2}N$, there exists $n_2\in N$ such that $b_1{n}_2=b_2$. Let $k\in a_2{b}_{2}N$. Then there exists $n\in N$ such that $k=a_2{b}_{2}n$. So $$\begin{array}{rcl}k& =& a_2{b}_{2}n\\ & =& a_1{n}_1{b}_1{n}_{2}n\\ & =& a_1{b}_1{b_1}^{-1}{n}_1{b}_1{n}_{2}n.\end{array}$$ Since $N$ is normal, ${b_1}^{-1}{n}_1{b}_1\in N$, and therefore $\left({b_1}^{-1}{n}_1{b}_1\right)n_{2}n\in N$. So $k=a_1{b}_1\left({b_1}^{-1}{n}_1{b}_1\right)n_{2}n\in a_1{b}_{1}N$. So $a_2{b}_{2}N\subseteq a_1{b}_{1}N$. So $\left(a_1{b}_1\right)N=\left(a_2{b}_2\right)N$. So the operation is well defined. The coset $N=1N$ is the identity since if $g\in G$ then $$\begin{array}{rcl}\left(N\right)\left(gN\right)& =& \left(1G\right)N\\ & =& gN\\ & =& \left(g1\right)N\\ & =& \left(gN\right)\left(N\right),\end{array}$$ If $gN$ is a coset then its inverse is $g^{-1}N$ since $$\begin{array}{rcl}\left(gN\right)\left(g^{-1}N\right)& =& \left(g{g}^{-1}\right)N\\ & =& N\\ & =& g^{-1}gN\\ & =& \left(g^{-1}N\right)\left(gN\right).\end{array}$$ So $G/N$ is a group. $\Leftarrow$ Assume $G/N$ is a group with operation $\left(aN\right)\left(bN\right)=abN$. To show: If $g\in G$ and $n\in N$ then $gn{g}^{-1}\in N$. First we show: If $n\in N$ then $nN=N$. Assume $n\in N$. To show: a. $nN\in N$. $\phantom{\text{To show:}}$ b. $N\in nN.$ Assume $x\in nN$. Then there exists $m\in N$ such that $x=nm$. Since $N$ is a subgroup, $nm\in N$. So $x\in N$. So $nN\subseteq N$. Assume $m\in N$. Then, since $N$ is a subgroup, $m=n{n}^{-1}m\in nN$. So $N\subseteq nN$. Now let $g\in G$ and $n\in N$. Then, by definition of the operation, $$\begin{array}{cc}gn{g}^{-1}N& =& \left(gN\right)\left(nN\right)\left(g^{-1}N\right)\\ & =& \left(gN\right)\left(N\right)\left(g^{-1}N\right)\\ & =& g1g^{-1}N\\ & =& N.\end{array}$$ So $gn{g}^{-1}\in N$. So $N$ is a normal subgroup of $G$. $\square$

• The quotient group $G/N$ is the set of cosets of the normal subgroup $N$ of the group $G$ with operation given by $\left(aN\right)\left(bN\right)=\left(abN\right)$.

Wow!! We actually made this weird set of cosets into a group!!

Let $N$ be a subgroup of a group $G$. Then $N$ is a normal subgroup of $G$ if and only if $G/N$ with operation given by $\left(aN\right)\left(bN\right)=\left(abN\right)$ is well defined.
HW: Show that if $G=N$ then $G/N\simeq \left\{1\right\}$.

Kernel and image of a homomorphism

• The kernel of a group homomorphism $f:G\to H$ is the set $$\ker f=\{g\in G|f\left(g\right)=1_H\},$$ where $1_H$ is the identity in $H$.
• The image of a group homomorphism $f:G\to H$ is the set $$\mathrm{im}f=\left\{f\right(g)\in H|g\in G\}.$$

Let $f:G\to H$ be a group homomorphism. Then

(a)   $\ker f$ is a normal subgroup of $G$.

(b)   $\mathrm{im}f$ is a subgroup of $H$.

		Proof.
	To show: $\ker f$ is a normal subgroup of $G$. $\mathrm{im}f$ is a subgroup of $G$. To show: aa. $\ker f$ is a subgroup. $\phantom{\text{To show:}}$ ab. $\ker f$ is normal. To show: aaa. If $k_1,k_2\in \ker f$ then $k_1{k}_2\in \ker f$. $\phantom{\text{To show:}}$ aab. $1_G\in \ker f$. $\phantom{\text{To show:}}$ aac. If $k\in \ker f$ then $k^{-1}\in \ker f$. Assume $k_1,k_2\in \ker f$. Then $f\left(k_1\right)=1_H$ and $f\left(k_2\right)=1_H$. So $f\left(k_1{k}_2\right)=f\left(k_1\right)f\left(k_2\right)=1_H$. So $k_1{k}_2\in \ker f$. Since $f\left(1_G\right)=1_H,1_G\in \ker f$. Assume $k\in \ker f$. So $f\left(k\right)=1_H$. Then $$f\left(k^{-1}\right)=f{\left(k\right)}^{-1}={1_H}^{-1}=1_H.$$ So $k^{-1}\in \ker f$. So $\ker f$ is a subgroup. To show: If $g\in G$ and $k\in \ker f$ then $gk{g}^{-1}\in \ker f$. Assume $g\in G$ and $k\in \ker f$. Then $$\begin{array}{ccc}f\left(gk{g}^{-1}\right)& =& f\left(g\right)f\left(k\right)f\left(g^{-1}\right)\\ & =& f\left(g\right)f\left(g^{-1}\right)\\ & =& f\left(g\right)f{\left(g\right)}^{-1}\\ & =& 1.\end{array}$$ So $gk{g}^{-1}\in \ker f$. So $\ker f$ is a normal subgroup of $G$. To show: $\mathrm{im}f$ is a subgroup of $H$. To show: ba. If $h_1,h_2\in \mathrm{im}f$ then $h_1{h}_2\in \mathrm{im}f$. $\phantom{\text{To show:}}$ bb. $1_H\in \mathrm{im}f$. $\phantom{\text{To show:}}$ bc. If $h\in \mathrm{im}f$ then $h^{-1}\in \mathrm{im}f.$ Assume $h_1,h_2\in \mathrm{im}f$. Then there exist $g_1,g_2\in G$ such that $h_1=f\left(g_1\right)$ and $h_2=f\left(g_2\right)$. Then $$h_1{h}_2=f\left(g_1\right)f\left(g_2\right)=f\left(g_1{g}_2\right),$$ since $f$ is a homomorphism. So $h_1{h}_2\in \mathrm{im}f$. By Proposition 2.1 (a), $f\left(1_G\right)=1_H$. So $1_H\in \mathrm{im}f$. Assume $h\in \mathrm{im}f$. Then there exists $g\in G$ such that $h=f\left(g\right)$. Then, by Proposition 2.1 (b), $$h^{-1}=f{\left(g\right)}^{-1}=f\left(g^{-1}\right).$$ So $h^{-1}\in \mathrm{im}f$. So $\mathrm{im}f$ is a subgroup of $H$. $\square$

Let $f:G\to H$ be a group homomorphism. Let $1_G$ be the identity element of $G$. Then

(a)   $\ker f=\left\{1_G\right\}$ if and only if $f$ is injective.

(b)   $\mathrm{im}f=H$ if and only if $f$ is surjective.

		Proof.
	Let $1_G$ and $1_H$ be the identities for $G$ and $H$, respectively. $\Rightarrow$ Assume $\ker f=\left(1_G\right)$. To show: If $f\left(g_1\right)=f\left(g_2\right)$ then $g_1=g_2$. Assume $f\left(g_1\right)=f\left(g_2\right)$. Then, by Proposition 2.1 (b), and the fact that $f$ is a homomorphism, $$1_H=f\left(g_1\right){f\left(g_2\right)}^{-1}=f\left(g_1{g_2}^{-1}\right).$$ So $g_1{g_2}^{-1}\in \ker f$. Since $\ker f=\left(1_G\right)$, then $g_1{g_2}^{-1}=1_G$. So $g_1=g_2$. So $f$ is injective. $\Leftarrow :$ Assume $f$ is injective. To show: aa. $\left(1_G\right)\subseteq \ker f$. $\phantom{\text{To show:}}$ ab. $\ker f\subseteq \left(1_G\right)$. Since $f\left(1_G\right)=1_H,1_G\in \ker f$. So $\left(1_G\right)\subseteq \ker f$. (Proof) Let $k\in \ker f$. Then $f\left(k\right)=1_H$. So $f\left(k\right)=f\left(1_G\right)$. Thus, since $f$ is injective, $k=1_G$. So $\ker f\subseteq \left(1_G\right)$. So $\ker f=\left(1_G\right)$. $\Rightarrow$ Assume $\mathrm{im}f=H$. To show: If $h\in H$ then there exists $g\in G$ such that $f\left(g\right)=h$. Assume $h\in H$. Then $h\in \mathrm{im}f$. So there exists $g\in G$ such that $f\left(g\right)=h$. So $f$ is surjective. $\Leftarrow$ Assume $f$ is surjective. To show: ba. $\mathrm{im}f\subseteq H$. $\phantom{\text{To show:}}$ bb. $H\subseteq \mathrm{im}f$. Let $x\in \mathrm{im}f$. Then there exists $g\in G$ such that $x=f\left(g\right)$. By the definition of $f,f\left(g\right)\in H$. So $x\in H$. So $\mathrm{im}f\subseteq H$. (Proof) Assume $x\in H$. Since $f$ is surjective there exists a $g$ such that $f\left(g\right)=x$. So $x\in \mathrm{im}f$. So $H\subseteq \mathrm{im}f$. So $\mathrm{im}f=H$. $\square$

Notice that the proof of Proposition (gpinjsur)(b) does not use the fact that $f:G\to H$ is a homomorphism, only the fact that $f:G\to H$ is a function.

HW: Show that if $S$ and $T$ are sets and $f:S\to T$ is a functions then $\mathrm{im}f=T$ if and only if $f$ is surjective.

(a)   Let $f:G\to H$ be a group homomorphism and let $K=\ker f$. Define $$\begin{array}{cccc}\hat{f}:& G/\ker f& ⟶& H\\ & gK& ⟼& f\left(g\right).\end{array}$$ Then $\hat{f}$ is a well defined injective group homomorphism.

(b)   Let $f:G\to H$ be a group homomorphism and define $$\begin{array}{cccc}f\prime :& G& ⟶& \mathrm{im}f\\ & g& ⟼& f\left(g\right).\end{array}$$ Then $f\prime$ is a well defined surjective group homomorphism.

(c)   If $f:G\to H$ is a group homomorphism then $$G/\ker f\simeq \mathrm{im}f,$$ where the isomorphism is a group isomorphism.

		Proof.
	To show: aa. $\hat{f}$ is well defined. $\phantom{\text{To show:}}$ ab. $\hat{f}$ is injective. $\phantom{\text{To show:}}$ ac. $\hat{f}$ is a homomorphism. To show: aaa. If $g\in G$ then $\hat{f}\left(gK\right)\in H$. $\phantom{\text{To show:}}$ aab. If $g_{1}K=g_{2}K$ then $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)$. Assume $g\in G$. Then $\hat{f}\left(gK\right)=f\left(g\right)$ and $f\left(g\right)\in H$ by the definition of $\hat{f}$ and $f$. Assume $g_{1}K=g_{2}K$. Then there exists $k\in K$ such that $g_1=g_{2}k$. To show: $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)$. To show: $f\left(g_1\right)=f\left(g_2\right)$. Since $k\in \ker f$, then $f\left(k\right)=1$ and so $$f\left(g_1\right)=f\left(g_{2}k\right)=f\left(g_2\right)f\left(k\right)=f\left(g_2\right).$$ So $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)$. So $\hat{f}$ is well defined. To show: If $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{1}K\right)$ then $g_{1}K=g_{2}K$. Assume $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)$. Then $f\left(g_1\right)=f\left(g_2\right)$. So $f\left(g_1\right)f{\left(g_2\right)}^{-1}=1$. So $f\left(g_1{g_2}^{-1}\right)=1$. So $g_1{g_2}^{-1}\in \ker f$. So $g_1{g_2}^{-1}=k$ for some $k\in \ker f$. So $g_1=g_{2}k$ for some $k\in \ker f$. To show: aba. $g_{1}K\subseteq g_{2}K$. $\phantom{\text{To show:}}$ abb. $g_{2}K\subseteq g_{1}K$. Let $g\in g_{1}K$. Then there exists $k_1\in K$ such that $g=g_1{k}_1$. So $g=g_{2}k{k}_1\in g_{2}K,$ since $k{k}_1\in K$. So $g_{1}K\subseteq g_{2}K$. Let $g\in g_{2}K$. Then $g=g_2{k}_2$ for some $k_2=k$. So $g=g_1{k}^{-1}{k}_2\in g_{1}K$, since $k^{-1}{k}_2\in K$. So $g_{2}K\subseteq g_{1}K$. So $g_{1}K=g_{2}K$. So $\hat{f}$ is injective. To show: $\hat{f}\left(g_{1}K\right)\hat{f}\left(g_{2}K\right)=\hat{f}\left(\right(g_{1}K\left)\right(g_{2}K\left)\right)$. Since $f$ is a homomorphism, $$\begin{array}{rcl}\hat{f}\left(g_{1}K\right)\hat{f}\left(g_{2}K\right)& =& f\left(g_1\right)f\left(g_2\right)\\ & =& f\left(g_1{g}_2\right)\\ & =& \hat{f}\left(g_1{g}_{2}K\right)\\ & =& \hat{f}\left(\right(g_{1}K\left)\right(g_{2}K\left)\right).\end{array}$$ So $\hat{f}$ is a homomorphism. To show: ba. $f\prime$ is well defined. $\phantom{\text{To show:}\phantom{}}$ bb. $f\prime$ is surjective. $\phantom{\text{To show:}\phantom{}}$ bc. $f\prime$ is a homomorphism. This is proved in Ex 2.2.3, Part I ?????????????? This is proved in Ex 2.2.3, Part I ?????????????? Since $f$ is a homomorphism, $$f\prime \left(g\right)f\prime \left(h\right)=f\left(g\right)f\left(h\right)=f\left(gh\right)=f\prime \left(gh\right).$$ So $f\text{'}$ is a homomorphism. Let $K=\ker f.$ By (a), the function $$\begin{array}{rcll}\hat{f}:& G/K& \to & H\\ & gK& \mapsto & f\left(g\right)\end{array}$$ is a well defined injective homomorphism. By (b), the function $$\begin{array}{rcll}\hat{f}\prime :& G/K& \to & \mathrm{im}\hat{f}\\ & gK& \mapsto & f^{^}\left(gK\right)=f\left(g\right)\end{array}$$ is a well defined surjective homomorphism. To show: ca. $\mathrm{im}\hat{f}=\mathrm{im}f$. $\phantom{\text{To show:}}$ cb. $\hat{f}\text{'}$ is injective. To show: caa. $\mathrm{im}\hat{f}\subseteq \mathrm{im}f$. $\phantom{\text{To show:}}$ cab. $\mathrm{im}f\subseteq \hat{f}$. Let $h\in \mathrm{im}\hat{f}$. Then there exists $gK\in G/K$ such that $\hat{f}\left(gK\right)=h$. Let $g\prime \in gK$. Then there exists $k\in K$ such that $g\prime =gk$. Then, since $f$ is a homomorphism and $f\left(k\right)=1$, $$\begin{array}{rcl}f(g\prime )& =& f\left(gk\right)\\ & =& f\left(g\right)f\left(k\right)\\ & =& f\left(g\right)\\ & =& \hat{f}\left(gK\right)\\ & =& h.\end{array}$$ So $h\in \mathrm{im}f$. So $\mathrm{im}\hat{f}\subseteq \mathrm{im}f$. Let $h\in \mathrm{im}f$. Then there exists $g\in G$ such that $f\left(g\right)=h$. So $\hat{f}\left(gK\right)=f\left(g\right)=h$. So $h\in \mathrm{im}\hat{f}$. So $\mathrm{im}f\subseteq \mathrm{im}\hat{f}$. To show: If $\hat{f}\text{'}\left(g_{1}K\right)=\hat{f}\text{'}\left(g_{2}K\right)$ then $g_{1}K=g_{2}K$. Assume $\hat{f}\prime \left(g_{1}K\right)=\hat{f}\prime \left(g_{2}K\right)$. Then $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)$. Then, since $\hat{f}$ is injective, $g_{1}K=g_{2}K$. So $\hat{f}\prime$ is injective. Thus $$\begin{array}{rcll}\hat{f}\prime :& G/K& \to & \mathrm{im}\hat{f}\\ & gK& \mapsto & f\left(g\right)\end{array}$$ is a well defined bijective homomorphism. $\square$

Direct products

Suppose $H$ and $K$ are groups. The idea is to make $H\times K$ into a group.

• The direct product $H\times K$ of two groups $H$ and $K$ is the set $H\times K$ with the operation given by $$(h_1,k_1)(h_2,k_2)=(h_1{h}_2,k_1{k}_2),$$ for $h_1,h_2\in H$, $k_1,k_2\in K$.
• More generally, given groups $G_1,G_2,\dots ,G_{\mathrm{n}}$, the direct product $G_1\times G_2\times \cdots \times G_n$ is the set $G_1\times G_2\times \cdots \times G_n$ with the operations given by $$(h_1,\dots ,h_i,\dots ,h_n)+(k_1,\dots ,k_i,\dots ,k_n)=(h_1{k}_1,\dots ,h_i{k}_i,\dots ,h_n{k}_n)$$ where $h_i,k_i\in G_i$, and $h_i{k}_i$ is given by the operation in the group $G_i$. The operation in the direct product is just the operations from the original groups acting componentwise.

HW: Show that these are good definitions, i.e. that, as defined above, $H\times K$ and $G_1\times G_2\times \cdots \times G_n$ are groups with identities given by $(1_H,1_K)$ and $(1_{G_1},\dots ,1_{G_n})$, respectively ($1_{G_i}$ denotes the identity in the group $G_i$).

Further Definitions

• An abelian group is a group $G$ such that if $g_1,g_2\in G$ then $g_1{g}_2=g_2{g}_1$.
• The center $Z\left(G\right)$ of a group $G$ is the set $$Z\left(G\right)=\left\{c\in G\right|\text{if}g\in G\text{then}cg=gc\}.$$

HW: Give an example of a non-abelian group.

HW: Prove that every subgroup of an abelian group is normal.

HW: Prove that $Z\left(G\right)$ is a normal subgroup of $G$.

HW: Prove that $Z\left(G\right)=G$ if and only if $G$ is abelian.

• The order $\left|G\right|$ of a group $G$ is the number of elements in $G$.
• Let $G$ be a group and let $g\in G$. The order $o\left(g\right)$ of $g$ is the smallest positive integer $n$ such that $g^n=1$. If no such integer exists then $o\left(g\right)=\infty$.

• Let $G$ be a group and let $S$ be a subset of $G$. The subgroup generated by $S$ is the subgroup $⟨S⟩$ of $G$ such that

  (a)   $S\subseteq $ ⟨S⟩$,$

  (b)   If $H$ is a subgroup of $G$ and $S\subseteq H$ then $$ ⟨S⟩\subseteq H$.$

$⟨S⟩$ is the smallest subgroup of $G$ containing $S$. Think of $⟨S⟩$ as gotten by adding to $S$ exactly those elements of $G$ that are needed to make a group.

HW: Let $G$ be a group and let $S$ be a subset of $G$. Show that the subgroup generated by $S$ exists and is unique.

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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